In right triangle $ABC$, $AB=9$, $BC=13$, and $\angle B = 90^\circ$.  Points $D$ and $E$ are midpoints of $\overline{AB}$ and $\overline{AC}$ respectively; $\overline{CD}$ and $\overline{BE}$ intersect at point $X$.  Compute the ratio of the area of quadrilateral $AEXD$ to the area of triangle $BXC$.
We begin by drawing a diagram: [asy]
pair A,B,C,D,E,X;
A=(0,9); B=(0,0); C=(13,0); E=(A+C)/2; D=(A+B)/2; X = intersectionpoint(B--E,D--C); label("$X$",X,N);
fill(A--E--X--D--cycle,rgb(135,206,250));

fill(B--X--C--cycle,rgb(107,142,35));
draw(A--B--C--cycle);
draw(C--D); draw(B--E);
draw(rightanglemark(A,B,C,15));
label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,NE);

label("$13$",(6.5,0),S); label("$9$",(-2,4.5),W);

draw((-2.7,5.3)--(-2.7,9),EndArrow(TeXHead));draw((-2.7,3.7)--(-2.7,0),EndArrow(TeXHead));
[/asy]

Since $D$ and $E$ are midpoints, $\overline{CD}$ and $\overline{BE}$ are medians.  Let $F$ be the midpoint of $\overline{BC}$; we draw median $\overline{AF}$.  The medians of a triangle are always concurrent (pass through the same point), so $\overline{AF}$ passes through $X$ as well.

[asy]
pair A,B,C,D,E,X,F;
A=(0,9); B=(0,0); C=(13,0); E=(A+C)/2; D=(A+B)/2; X = intersectionpoint(B--E,D--C); label("$X$",X,N);

F=(B+C)/2; draw(A--F,dashed); label("$F$",F,S);

draw(A--B--C--cycle);
draw(C--D); draw(B--E);
draw(rightanglemark(A,B,C,15));
label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,NE);

[/asy]

The three medians cut triangle $ABC$ into six smaller triangles.  These six smaller triangles all have the same area.  (To see why, look at $\overline{BC}$ and notice that $\triangle BXF$ and $\triangle CXF$ have the same area since they share an altitude and have equal base lengths, and $\triangle ABF$ and $\triangle ACF$ have the same area for the same reason.  Thus, $\triangle ABX$ and $\triangle ACX$ have the same area.  We can repeat this argument with all three sizes of triangles built off the other two sides $\overline{AC}$ and $\overline{AB}$, to see that the six small triangles must all have the same area.)

Quadrilateral $AEXD$ is made up of two of these small triangles and triangle $BXC$ is made up of two of these small triangles as well.  Hence they have the same area (and this will hold true no matter what type of triangle $\triangle ABC$ is).  Thus, the ratio of the area of quadrilateral $AEXD$ to the area of triangle $BXC$ is $1/1=\boxed{1}$.